Question: Is ${945936}$ divisible by $9$ ?
Answer: A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {945936}= &&{9}\cdot100000+ \\&&{4}\cdot10000+ \\&&{5}\cdot1000+ \\&&{9}\cdot100+ \\&&{3}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {945936}= &&{9}(99999+1)+ \\&&{4}(9999+1)+ \\&&{5}(999+1)+ \\&&{9}(99+1)+ \\&&{3}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {945936}= &&\gray{9\cdot99999}+ \\&&\gray{4\cdot9999}+ \\&&\gray{5\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {9}+{4}+{5}+{9}+{3}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${945936}$ is divisible by $9$ if ${ 9}+{4}+{5}+{9}+{3}+{6}$ is divisible by $9$ Add the digits of ${945936}$ $ {9}+{4}+{5}+{9}+{3}+{6} = {36} $ If ${36}$ is divisible by $9$ , then ${945936}$ must also be divisible by $9$ ${36}$ is divisible by $9$, therefore ${945936}$ must also be divisible by $9$.